Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
A car driving at 45 km/h brakes with a constant deceleration of 4.5 m/s² until it stops. Find the braking distance s.
A ball is thrown straight up at 10 m/s from the edge of a roof 20 m above the ground. Ignore air resistance and use g = 9.81 m/s². How long after the throw does it hit the ground? Find t.
An 18 kg block slides down a slope at 30° to the horizontal. The coefficient of kinetic friction is 0.3. Use g = 9.81 m/s². Find the acceleration a of the block along the slope.
A 7 kg block on a horizontal table is connected by a light string over a frictionless pulley to a hanging 3.5 kg mass. The coefficient of kinetic friction between block and table is 0.3. Use g = 9.81 m/s². Find the acceleration a of the system.
A ball is launched from level ground at 35° above the horizontal and lands 105 m away. Ignore air resistance and use g = 9.81 m/s². Find the launch speed v0.
A ball is launched at 45° above horizontal from the top of a 15 m cliff. Ignore air resistance. Find the time of flight t until the ball hits the ground.
A 1000 kg car drives up a 3° slope at a constant 90 km/h. Air resistance and rolling resistance together amount to 450 N. Find the power P the wheels must deliver, in kW.
A 10 kg block is released from rest at height 3.5 m on a smooth ramp. At the bottom it crosses a rough horizontal strip 3 m long (μ = 0.15) and then runs onto a spring (k = 4000 N/m) on a smooth surface. Find the maximum compression x of the spring.
A 12 kg projectile moving horizontally at 29 m/s explodes into two fragments that move along the same line. The 10 kg fragment continues forwards at 35 m/s. Find the velocity v′2 of the 2 kg fragment (positive in the original direction).
A 0.5 kg ball hits a wall at 10 m/s and bounces straight back at 15 m/s. The contact lasts 25 ms. Find the magnitude of the impulse J the wall gives the ball.
A vertical rectangular gate 2 m wide sits in the wall of a tank of water, with its top edge 2 m below the surface. Find the resultant hydrostatic force F on the gate.
A vertical rectangular gate 3 m wide sits in the wall of a tank of seawater, with its top edge 0.5 m below the surface. Find the depth ycp of the centre of pressure below the surface.
A pitot-static tube in a water pipe reads a difference of 15 kPa between the stagnation pressure and the static pressure. Find the water velocity v at the tip.
A fountain nozzle points straight up. Just below the nozzle the supply pipe has a diameter of 75 mm and a gauge pressure of 130 kPa. The nozzle exit diameter is 20 mm, and the height difference across the nozzle is negligible. Ignoring losses and air resistance, find the height H the jet rises above the nozzle. Use g = 9.81 m/s².
What volume V does 310 g of He occupy at 500 kPa and 0 °C? Treat it as an ideal gas.
Find the density ρ of Ar at 250 kPa and 30 °C, treating it as an ideal gas.
A heat engine receives heat from a source at 500 °C and rejects heat to the surroundings at 15 °C. Its actual thermal efficiency is 35%. Find its second-law efficiency ηII, the ratio of the actual efficiency to the Carnot efficiency.
A heat pump keeps a house at 21 °C while it is 5 °C outside, delivering 13 kW of heat to the house. Find the minimum power input Ẇmin.
A string fixed at both ends is 0.75 m long; waves travel on it at 80 m/s. Find the frequency f3 of its harmonic number 3.
A diffraction grating has 1200 lines per millimetre. Light of wavelength 656 nm falls on it at right angles. What is the highest order mmax of bright maximum that can be seen?
Answer: s = 17.36 m
Motion with constant acceleration: v = v₀ + a·t, s = v₀·t + ½·a·t² and v² = v₀² + 2·a·s. Pick the equation that contains the three things you know and the one you want, and keep the signs consistent with one positive direction.
Answer: t = 3.281 s
Motion with constant acceleration: v = v₀ + a·t, s = v₀·t + ½·a·t² and v² = v₀² + 2·a·s. Pick the equation that contains the three things you know and the one you want, and keep the signs consistent with one positive direction.
Up is positive and the ground is h₀ below the starting point.
Only the positive root is physical.
Answer: a = 2.356 m/s²
Newton’s second law: the net force on a body equals its mass times its acceleration, ΣF = m·a. Draw a free-body diagram, choose a positive direction along the motion, and add every force in that direction: applied forces, friction μ·N against the motion, and the components of gravity.
Down the slope is positive.
Answer: a = 1.308 m/s²
Newton’s second law: the net force on a body equals its mass times its acceleration, ΣF = m·a. Draw a free-body diagram, choose a positive direction along the motion, and add every force in that direction: applied forces, friction μ·N against the motion, and the components of gravity.
Answer: v0 = 33.11 m/s
Projectile motion: with no air resistance, x = vxt and y = h + vyt − ½gt². The two directions are independent except that they share the same time.
Answer: t = 4.816 s
Projectile motion: with no air resistance, x = vxt and y = h + vyt − ½gt². The two directions are independent except that they share the same time.
Horizontal velocity stays constant; vertical motion has a constant acceleration g = 9.81 m/s² downward.
Solve y(t) = h + vy·t − ½gt² = 0 for the positive root.
Answer: P = 24.09 kW
Work–energy principle: the change in kinetic energy equals the work done by all forces. Writing gravity and springs as potential energy (mgh and ½kx²) turns this into a balance: initial energy = final energy + energy lost to friction (μN·d). Power is force times speed: P = F·v.
Answer: x = 38.68 cm
Work–energy principle: the change in kinetic energy equals the work done by all forces. Writing gravity and springs as potential energy (mgh and ½kx²) turns this into a balance: initial energy = final energy + energy lost to friction (μN·d).
The ramp is smooth, so only the strip removes energy.
Answer: v′2 = −1 m/s
Conservation of momentum: when no external force acts during a collision, the total momentum Σm·v before equals the total after. Velocities have a direction, so pick a positive direction and give velocities the other way a minus sign. In an elastic collision kinetic energy is conserved too; in a perfectly inelastic one the bodies stick together.
The minus sign means the second fragment flies backwards.
Answer: J = 12.5 N·s
Conservation of momentum: when no external force acts during a collision, the total momentum Σm·v before equals the total after. Velocities have a direction, so pick a positive direction and give velocities the other way a minus sign. In an elastic collision kinetic energy is conserved too; in a perfectly inelastic one the bodies stick together.
Answer: F = 206 kN
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Answer: ycp = 2.048 m
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Pressure grows with depth, so the resultant acts below the centroid.
Answer: v = 5.477 m/s
Continuity and Bernoulli: for an incompressible fluid the flow rate is the same at every section, Q = A₁v₁ = A₂v₂. Along a streamline without losses, p + ½ρv² + ρgz is constant, so where the fluid is faster or higher its pressure is lower.
The flow comes to rest at the tip, so its kinetic energy shows up as extra pressure.
Answer: H = 13.32 m
Continuity and Bernoulli: for an incompressible fluid the flow rate is the same at every section, Q = A₁v₁ = A₂v₂. Along a streamline without losses, p + ½ρv² + ρgz is constant, so where the fluid is faster or higher its pressure is lower.
Bernoulli along the jet: at the top the velocity is zero and the pressure is atmospheric again.
Answer: V = 351.7 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 3.963 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: ηII = 55.79 %
Second-law limits: no cycle working between reservoirs at absolute temperatures TH and TL can beat the reversible (Carnot) cycle: ηmax = 1 − TL/TH, COPR,max = TL/(TH − TL) and COPHP,max = TH/(TH − TL). Energy is always conserved: QH = W + QL.
Answer: Ẇmin = 0.7071 kW
Second-law limits: no cycle working between reservoirs at absolute temperatures TH and TL can beat the reversible (Carnot) cycle: ηmax = 1 − TL/TH, COPR,max = TL/(TH − TL) and COPHP,max = TH/(TH − TL). Energy is always conserved: QH = W + QL.
The least work is needed by a reversible machine, which has the highest COP.
Answer: f3 = 160 Hz
Standing waves and the Doppler effect: a string fixed at both ends, or a pipe open at both ends, fits whole half-wavelengths: L = nλ/2, fn = n·v/(2L), n = 1, 2, 3, … A pipe closed at one end has a node at the closed end and an antinode at the open end: L = nλ/4 with odd n only, fn = n·v/(4L). When source or observer move along the line between them, f′ = f·(v ± vo)/(v ∓ vs): use the upper signs for motion towards each other. Two waves of nearly equal frequency beat at |f₁ − f₂|.
Answer: mmax = 1
Interference and diffraction: light from two slits a distance d apart reaches a point on the screen with a path difference d·sin θ. Bright fringes appear where d·sin θ = mλ (m = 0, ±1, ±2, …). For a screen at distance L ≫ d, the m-th bright fringe is at y = mλL/d and the fringe spacing is Δy = λL/d. A grating with N lines per millimetre has d = 1/N and sharp maxima at d·sin θ = mλ; since sin θ ≤ 1, the highest order is the largest whole number m ≤ d/λ. A single slit of width a has dark fringes at a·sin θ = mλ, m = ±1, ±2, …
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